Maximum Parsimony Scores: why the "highest score" requires the "fewest changes"

Informative positions of types 5, 6, & 7 favor the first, second, and third hypotheses of relationship, respectively, because each of these types requires a single nucleotide change in that hypothesis, and two changes in the alternative hypotheses.

Then, in a hypothetical data set [not shown] with 10 Type 5 positions, 5 Type 6 positions, and 2 Type 7 positions, 
    Hypothesis #1 has the highest score ("10") and requires N = 10 + 2(  5 + 2) = 24 changes
    Hypothesis #2 would require:                                            N =   5 + 2(10 + 2) = 29 changes
    Hypothesis #3 would require:                                            N =   2 + 2(10 + 5) = 32 changes

Thus, Hypothesis #1, which has the "highest score" in terms of the number of parsimony sites that favor it, requires the fewest changes and is therefore the most parsimonious explanation of the data.

Uninformative positions of types 1, 2, 3, & 4 require 0, 1, 3, & 2 nucleotide changes, respectively, in any of the hypotheses: the relative ranking of hypotheses is affected only by the informative sites.

Homework: SHOW that positions of types 1, 2, 3, & 4 require 0, 1, 3, & 2 nucleotide changes, respectively, for any of the three hypotheses above. [Hint: Draw the three hypotheses, and find at least one assignment of changes to the branches or internode that satisfies the statement.